Question: Let $x,$ $y,$ $z$ be positive real numbers such that $xyz = 8.$  Find the minimum value of $x + 2y + 4z.$
Answer: By AM-GM,
\[x + 2y + 4z \ge 3 \sqrt[3]{(x)(2y)(4z)} = 3 \sqrt[3]{8xyz} = 3 \sqrt[3]{8 \cdot 8} = 12.\]Equality occurs when $x = 2y = 4z$ and $xyz = 8.$  We can solve to get $x = 4,$ $y = 2,$ and $z = 1,$ so the minimum value is $\boxed{12}.$